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Maximal product of primes whose sum is bounded
Marc Deléglise, Jean-Louis Nicolas
To cite this version:
Maximal product of primes whose sum is bounded
Marc Deléglise and Jean-Louis Nicolas ∗September 17, 2012
To the memory of A.A. Karatsuba, on the occasion of his 75th birthday.
Abstract
Ifn is a positive integer, let h(n) denote the maximal value of the product q1q2. . . qj for all families of primes q1 < q2 < . . . < qj such that q1 + q2+ . . . + qj ≤ n. We shall give some properties of this functionh and describe an algorithm able to compute h(n) for any n up to1035
.
1
Introduction
1.1 Function h(n)
If n≥ 2 is an integer, let us define h(n) as the greatest product of a family of primes q1 < q2 < . . . < qj the sum of which does not exceed n.
Let ℓ be the additive function such that ℓ(pα) = pα for p prime and α ≥ 1. In other words, if the standard factorization of M into primes is M = qα1 1 qα22. . . q αj j , we have ℓ(M ) = q1α1+ qα22+ . . . + q αj j and ℓ(1) = 0. Note that ℓ(n) is sequence A008475 in OEIS (cf. [12]). If µ denotes the Möbius function, h(n) can also be defined as
(1.1) h(n) = max ℓ(M )≤n µ(M )6=0 M. Note that (1.2) ℓ(h(n))≤ n.
From the uniqueness of the factorization of h(n) into primes, the maximum in (1.1) is attained at only one point. It is convenient to set
h(0) = h(1) = 1.
∗
(h(n))n≥1is sequence A159685 of [12]. A table of the 50 first values of h(n) is given at the end of the paper. A larger table may be found on the authors’s web sites [2, 10].
In [9], Landau has introduced the function g(n) as the maximal order of an element in the symmetric group Sn; he has shown that
(1.3) g(n) = max
ℓ(M )≤nM.
The introductions of [5] and [3] recall the main properties of Landau’s func-tion g(n) which is menfunc-tioned as entry A002809 in [12]. From (1.1) and (1.3), it follows that
(1.4) h(n)≤ g(n), (n ≥ 0).
In this article, we shall give some properties of h(n) and describe an algorithm able to calculate h(n) for any n up to 1035.
1.2 Notation
• We denote by N the set of non-negative integers. • The symbol p will always denote a prime number. • For every arithmetic function f : N → C, we define
(1.5) πf(x) =
X p≤x, p prime
f (p)
• In particular, for f(n) = 1, we will note, as usual π(x) = π1(x) the number of primes up to x. • For f(n) = n we define (1.6) πid(x) = X p≤x, p prime p
• We denote by pj the j-th prime and we set σ0 = 0, N0 = 1 and, for j≥ 1,
(1.7) σj = πid(pj) = p1+ p2+ . . . + pj, Nj = p1p2. . . pj. In § 3, for all j≥ 1, we shall prove that h(σj) = Nj.
• If m is an integer, we denote by (m)⋆ the smallest prime p satisfying p≥ m and, if m ≥ 2, by⋆(m) the largest prime p satisfying p≤ m. • P+(m) (resp. P−(m)) will denote the largest (resp. smallest) prime
• ω(n) is the number of distinct prime factors of n and Ω(n) the num-ber of prime factors of n, counted with multiplicity. µ(n) is Möbius’s function.
• for x > 1, we denote by log x the natural logarithm of x and by log log x = log(log x) the iterated logarithm of x.
• Li is the integral logarithm defined for x > 1 by Li(x) = lim ε→0, ε>0 Z 1−ε 0 + Z x 1+ε dt
log t = γ + log log x + X n≥1
(log x)n nn! where γ is Euler’s constant.
1.3 Functions hj(n)
For n≥ 0, let k = k(n) be the non-negative integer defined by (1.8) σk = πid(pk)≤ n < πid(pk+1) = σk+1.
It is the maximal number of prime factors of h(n). For 0 ≤ j ≤ k = k(n), let us set
(1.9) hj(n) = max
ℓ(M )≤n µ(M )6=0, ω(M)=j
M
where ω(M ) is the number of prime factors of M . For n≥ 0, we have
(1.10) h0(n) = 1
while, for n≥ 2, we have
(1.11) h1(n) =⋆(n)≥ 2.
Note that
(1.12) ℓ(hj(n))≤ n.
In § 6, we prove that, for all n’s, the sequence hj(n) is increasing on j, so that
(1.13) h(n) = hk(n), (n≥ 0).
Our proof is not that simple. A possible reason is that this increasingness relies on the properties of the whole set of primes P. Let P′ be a subset of P and NP′ the set of integers whose prime factors belong to P′. We may
consider
(1.14) hj(n,P′) = max
M∈NP′, ℓ(M )≤n
By choosingP′ ={2, 3, 11, 13, 17, 19, 23, . . .} = P \ {5, 7}, we observe that h2(24,P′) = 11· 13 = 143 > h3(24,P′) = 2· 3 · 19 = 114.
In § 4, we give an upper bound for hj(n) which will be useful in § 6 where our proof of the increasingness of hj is given. In (1.9), hj(n) can be considered as the solution of a problem of optimization with prime variables. The upper bound of hj(n) is obtained by relaxing some constraints so that certain variables are no longer primes, but only integers.
1.4 Elementary computation of h(n) and hj(n)
The naive algorithm described in [5] to compute g(n) can be easily adapted to calculate h(n) for 1≤ n ≤ N. Note that, for the prime factors of h(n), Corollary 3.1 below furnishes the upper bound
P+(h(n))≤ pk(n)+1+ pk(n)+2.
It also can be adapted to compute hj(n). For r ≥ j ≥ 1 and n ≥ σj, let us define
h(r)j (n) = max P+(M )≤pr, ℓ(M )≤n
µ(M )6=0, ω(M)=j M . We have the induction relation
h(r+1)j (n) = max(h(r)j (n), pr+1h(r)j−1(n− pr+1)). Indeed, either pr+1 does not divide h(r+1)j (n), and h
(r+1) j (n) = h (r) j (n) holds, or pr+1divides h(r+1)j (n), and h (r+1) j (n) = pr+1h(r)j−1(n−pr+1), which implies n≥ pr+1+ σj−1.
Moreover, if pr ≥ n, we have h(r)j (n) = hj(n), h(r)r (n) = Nrand h(r)1 (n) = ⋆(n) for n < p
r while, for n ≥ pr, h(r)1 (n) = pr holds. So, we may write algorithm 1, which has been used to calculate the table in appendix. The merging and pruning method described in [5, §2.2] can be used to improve the running time.
In § 8, a more sophisticated algorithm to calculate h(n) is given. It is based on a fast method to compute πid(x), which is explained in § 7.
2
Some lemmas
Lemma 2.1. If m ≥ 2 is an integer, let us denote by (m)⋆ (resp. ⋆(m)) the smallest (resp. largest) prime p satisfying p ≥ m (resp. p ≤ m). Then
(m)⋆ ≤ 11
8 m and
Algorithm 1Computation of hj(n) for 2≤ n ≤ nmax and 1 ≤ j ≤ k(n) Procedure ComputeHj(nmax)
r = 1; p = pr; kmax = k(nmax); pmax = pkmax+1+ pkmax+2 whilep≤ pmax do
forn from σr to nmax do H[r, n] = Nr
jmax = min(r− 1, kmax) forj from jmax by−1 to 2 do
forn from nmax by −1 to p + σj−1 do H[j, n] = max(H[j, n], p∗ H[j − 1, n − p]) forn from p to nmax do
H[1, n] = p; r = r + 1; p = pr
Proof. We use the result of [6]: for x ≥ 396738, the interval [x, x + x 25 log2x]
contains a prime number. As 396833 is prime, we deduce that, for pi ≥ 396833, (2.1) pi+1 pi ≤ 1 + 1 25 log2pi ≤ 1 + 1 25 log2396833 < 1.00025 < 11 8 < 10 7 . If m is prime, (m)⋆ =⋆(m) = m holds, while, if m is not prime, we define pi by pi < m < pi+1; we have (m)⋆ = pi+1,⋆(m) = pi,
(m)⋆ m ≤ pi+1 pi+ 1 < pi+1 pi , ⋆(m) m ≥ pi pi+1− 1 > pi pi+1
and, if pi ≥ 396833, the result follows from (2.1). Finally, it remains to check that pi+1 pi+ 1 ≤ 11 8 and pi pi+1− 1 ≥ 7 10 hold for all pi’s satisfying 2≤ pi< 396833.
Lemma 2.2. Let p < p′ be two primes. There exists a third prime p′′ satisfying
(2.2) p + p′ ≤ p′′≤ pp′− p + 1.
Proof. Let us show that p′′ = (p + p′)⋆ satisfies (2.2). By Lemma 2.1, it suffices to prove that 118(p + p′)≤ pp′− p + 1, i.e:
(2.3) pp′ 8−11 p − 19 p′ + 8 pp′ ≥ 0. If p≥ 3 and p′≥ 5, we have 11
p +19p′ ≤ 113 +195 < 8 and (2.3) holds. Similarly,
if p = 2 and p′ ≥ 11, the inequality 11p +19p′ ≤ 112 + 1911 < 8 implies (2.3). In
Lemma 2.3. Let p and p′ be two prime numbers satisfying 3 ≤ p < p′ and pp′ 6= 15. There exists a prime p′′ such that
(2.4) p + p′≤ p′′≤ 5
6pp ′− p.
Proof. The proof is similar to the one of the preceding lemma. From Lemma 2.1, to show that p′′ = (p + p′)⋆ satisfies (2.4), it suffices to show that
11
8(p + p′)≤ 56pp′− p, i.e. 33/p + 57/p′ ≤ 20, which evidently holds for p ≥ 3 and p′≥ 7.
Lemma 2.4. For all i ≥ 2, the following inequality
(2.5) pi+ pi−1 ≤ p2i−1
holds. Moreover, let b be a positive integer; there exists a positive integer i0 = i0(b) such that we have
(2.6) pi+ pi−1 < p2i−b for i ≥ i0(b). The table below gives some values of i0(b)
b = 1 2 3 4 5 6 7 8 9 10 12 13 18 30 3675
i0 = 3 4 7 8 18 19 27 28 36 39 50 53 85 149 33127
Proof. We start from the two inequalities
(2.7) pi ≤ i(log i + log log i − α), (α = 0.9484, i ≥ 39017),
(2.8) pi≥ i(log i + log log i − 1), (i ≥ 2) which can be found in [7]. From (2.7), it follows that
(2.9) pi−1+ pi≤ (2i − 1)(log i + log log i − α), (i ≥ 39018) while, if i≥ max(2, b), which implies 2i − b ≥ 2 and i ≥ b, (2.8) gives (2.10) p2i−b≥ (2i − b)(log i + log 2 + log
2i− b
2i
+ log log i− 1). By using the inequality log t≤ t − 1, we get
log 2i− b 2i =− log 2i 2i− b ≥ − 2i 2i− b − 1 =− b 2i− b and (2.10) yields
Under the condition
(2.12) i≥ max(39018, b),
the substraction of (2.9) from (2.11) gives
p2i−b− pi−1− pi ≥ (log i + log log i + log 2)(1 − b) + 2i(log 2− 1 + α) − log 2 − α > (log i + log log i + log 2)
1.283 i− 1.642
log i + log log i + log 2− (b − 1)
. (2.13)
Now, the two functions t7→ t/(log t + log log t + log 2) and t 7→ −1/(log t + log log t + log 2) are increasing for t≥ e2; choosing i1 = 39018 and
(2.14) b =
1 + 1.283 i1− 1.642 log i1+ log log i1+ log 2
=⌊3675.52 . . .⌋ = 3675 shows that, for i ≥ i1, (2.12) is satisfied and that in (2.13), the bracket is positive. Therefore, (2.13) proves pi+ pi−1< p2i−3675 for i≥ i1= 39018.
To determine the entries of the table, for all i’s up to 39018, we have calculated bi= 2i− 1 − π(pi+ pi−1) which is the smallest integer such that pi−1+ pi< p2i−bi. Further, for each b in the table, we have determined i0(b)
which is the smallest integer i0 such that, for i0(b)≤ i ≤ 39018, bi≥ b holds. As i0(1) = 3, for all i≥ 3, pi+ pi−1< p2i−1 holds. So, (2.5) follows from p2+ p1 = 3 + 2 = 5 = p3.
Lemma 2.5. Under Riemann hypothesis, for all x ≥ 41 we have (2.15) πid(x)− Li(x2)
≤ 24π5 x3/2log x.
Proof. Let us define r(x) by π(x) = Li(x) + r(x) and assume the Riemann hypothesis. Then cf. [11, (6.18)] :
(2.16) |r(x)| = |π(x) − Li(x)| ≤ 1 8π
√
x log x (for x≥ 2657). Let us denote x0 = 2657. Then, from (1.6), Stieltjes’s integral gives :
and, usingZ √t log t = 2 3t 3/2log t−2 3 , (2.17) πid(x)− Li(x2) ≤ 5 24πx 3/2log x− 1 18πx 3/2 + πid(x0)− Li(x20)− x0r(x0) −12π1 x3/20 log x0+ 1 18πx 3/2 0 . The computation of r(x0) = π(x0)− Li(x0) = 384− 399.59681 . . . = −15.59681 . . . πid(x0)− Li(x20) = 464 653− 480610.2863 . . . = −15957.2863 . . .
and (2.17) imply for x≥ x0, πid(x)− Li(x2) ≤ 24π5 x3/2log x− 1 18πx 3/2− 740.023 . . . ≤ 5 24πx 3/2log x.
which proves (2.15) for x ≥ x0 = 2657. It remains to check (2.15) for 41≤ x ≤ 2657; by setting f1(x) = Li(x2)− 5 24πx 3/2log x, f 2(x) = Li(x2) + 5 24πx 3/2log x, it is equivalent to check (2.18) f1(x)≤ πid(x)≤ f2(x)
for 41 ≤ x ≤ 2657. One remarks that f1 and f2 are increasing for x ≥ 2. Therefore, to prove (2.18), it suffices to check that for every prime p satisfying 41 ≤ p ≤ 2657 we have f1(p′) ≤ πid(x) ≤ f2(p) where p′ is the prime following p.
Note that, in the range [2..2657], πid(x)− Li(x2) has several changes of sign, the smallest one being for x = 110.35 . . .
Lemma 2.6. Let z and u be two real numbers satisfying z ≥ 1 and √z ≤ u ≤ z. Suppose that we have precomputed the tables prime, piftab and pi. The first two tables are indexed by the integers k, 0 ≤ k ≤ π(u), and the third one by the integers t, 0 ≤ t ≤ u.
Then the sum (2.19) X √ z<q≤u, q prime f (q)πf z q
may be computed in O(√z/ log z) time.
Proof. For q > √z, z/q belongs to [1, √z) . The number of primes in this interval is O(√z/ log z), thus the number of values of πf(z/q) is O(√z/ log z). We group the q’s for which πf(z/q) takes the same value. Algorithm 2 carries out this computation.
Algorithm 2: Computation of the sum (2.19) in O √z/ log z time S = 0; imin = 1 + pi[⌊√z⌋]
whileimin≤ pi[u] do q = prime[imin] s = pi[z/q]
imax = min(pi(z/ prime[s]), pi[u])
S = S + (piftab[imax]− piftab[imin − 1]) ∗ piftab[s] imin = imax + 1
return S
Let us give some words to convince of the correctness of algorithm 2 : let us note s = π(z/q). Then psis the largest prime≤ z/q. For q′prime, q′ ≥ q, we have πf(z/q′) = πf(ps) = piftab[s] if and only if z/q′ ≥ ps i.e. q′≤ z/ps, in other terms, π(q′)≤ π(z/ps). Thus the largest prime q′ in the range [q..u] such that πf(z/q′) = πf(ps) is pi where i = min(π(z/ps), π(u)).
3
First results
Proposition 3.1. Let j be a positive integer and σj and Nj be defined by (1.7). We have
h(σj) = Nj.
Proof. It is easy to see that h(σ1) = h(2) = 2 = N1 and h(σ2) = h(5) = 6 = N2. Now, we may suppose that j≥ 3, i.e. pj ≥ 5 and we set ρ = pj/ log pj. The function t7→ t/ log t is increasing for t ≥ e and, since 2/ log 2 < 5/ log 5, we have, for 1≤ i < j, pi/ log pi < ρ and for i > j, pi/ log pi > ρ; in other words, i− j and pi/ log pi− ρ have the same sign.
Let M be a product of r distinct primes, M = Q1Q2. . . Qr, with r≥ 0. After a possible simplification by s primes (0≤ s ≤ min(j, r)), we may write
M Nj
= pj1pj2. . . pju
with u = r− s, v = j − s and
pk1 < pk2 < . . . < pkv ≤ pj < pj1 < pj2 < . . . < pju.
Let f (M ) = ℓ(M )− ρ log M. From the definition of ℓ, the function f is additive and we have
(3.1) f (M )− f(Nj) = u X i=1 (pji− ρ log pji)− v X i=1 (pki− ρ log pki)≥ 0
since each term of the first sum is non-negative while, in the second sum, each term is non-positive.
From (1.1), since ℓ(Nj) = σj, in order to prove that h(σj) = Nj, we must show that, for all squarefree number M satisfying ℓ(M )≤ σj = ℓ(Nj), we have M ≤ Nj. But, for such an M , (3.1) yields
f (M ) = ℓ(M )− ρ log M ≥ f(Nj) = ℓ(Nj)− ρ log Nj = σj− ρ log Nj whence M Nj ≤ exp ℓ(M )− σj ρ ≤ 1, which completes the proof of Proposition 3.1.
Proposition 3.2. Let r and j be two positive integers and σj, Nj and hj be defined by (1.7) and (1.9). We have
(3.2) hj(σj+r− σr) = Nj+r/Nr = pr+1pr+2. . . pr+j. Moreover, if n ≥ σj+r− σr we have
(3.3) ℓ(hj(n))≥ σj+r− σr.
Proof. The proof is similar to the one of Proposition 3.1. Let us set (3.4) ρ = pj+r− pr
log(pj+r/pr)
and ρ′ = ρ log pr− pr.
Since, for t6= 1, (t − 1)/t < log t < t − 1 holds, we have pj+r> ρ > pr≥ 2. For a squarefree number M , we consider the additive function
f (M ) = ℓ(M )− ρ log M + ρ′ω(M ) =X p|M
f (P ) =X p|M
(p− ρ log p + ρ′). We will prove that f attains its minimum in N = Nj+r/Nr. From (3.4), it follows that f (pj+r) = f (pr) = 0 and the study of the function t 7→ t− ρ log t + ρ′ shows that
Therefore, we have (3.5) f (M )− f(N) = X p|M p < pr orp > pj+r f (p)− X p∤M pr< p < pj+r f (p)≥ 0.
From (1.9), we have to show that, for any squarefree integer M satisfying ℓ(M ) ≤ σj+r− σr = ℓ(N ) and ω(M ) = j = ω(N ), we have M ≤ N. For such an M , (3.5) gives ℓ(M )− ρ log M + ρ′ω(M )≥ ℓ(N) − ρ log N + ρ′ω(N ) yielding M N ≤ exp ℓ(M )− ℓ(N) ρ ≤ 1, which, together with ℓ(N ) = σj+r− σr, proves (3.2).
To prove (3.3), first, from (1.9), we observe that hj(n)≥ N = Nj+r/Nr. Setting M = hj(n) in (3.5) and noting that ω(M ) = ω(N ) = j, we see that
ℓ(M )≥ ℓ(N) + ρ logM
N ≥ ℓ(N) = σj+r− σr which proves (3.3).
Proposition 3.3. Let n ≥ 2 be an integer and p < p′ two prime numbers which do not divide h(n). Then the largest prime divisor P+(h(n)) of h(n) satisfies
P+(h(n)) < p + p′.
Proof. Let us assume that the set of prime factors of h(n) not smaller than p + p′ is not empty and let q≥ p + p′ be its smallest element.
• If q < pp′, by setting M = pp′
q h(n), we have by (1.2) ℓ(M ) = p + p′− q + ℓ(h(n)) ≤ ℓ(h(n)) ≤ n and thus, from (1.1),
(3.6) h(n)≥ M = pp′
q h(n), in contradiction with q < pp′.
We have p≥ 2, p′ ≥ 3 and p′′≥ p+p′≥ 5, so that 10
p +p7′′ ≤ 102 +75 < 7,
yielding q < pp′′. By considering M = ppq′′h(n), as in (3.6), we get h(n)≥ M = pp′′
q h(n) > h(n), a contradiction.
Corollary 3.1. If k = k(n) is defined by (1.8), the largest prime factor of h(n) satisfies
P+(h(n)) < pk+1+ pk+2.
Proof. The number of prime factors of h(n) does not exceed k, so that, among p1, p2, . . . , pk+2 there are certainly two prime numbers p and p′ not dividing h(n). By applying Proposition 3.3, we get P+(h(n)) < p + p′ ≤ pk+1+ pk+2.
Proposition 3.4. Let n ≥ 5 be an integer, k ≥ 2 be defined by (1.8) and j an integer satisfying 2 ≤ j ≤ k. Let us supose that there exists two prime numbers, p, p′ not dividing h
j−1(n), and satisfying 3 ≤ p < p′ and P+(h
j−1(n)) ≥ p + p′ where P+(hj−1(n)) is the largest prime divisor of hj−1(n). Then the inequality
hj(n) > 6
5hj−1(n) holds.
Proof. Let us consider two cases :
Case 1 : pp′ > 15. Let us denote by q≤ P+(h
j−1(n)) the smallest prime dividing hj−1(n) and satisfying p + p′ ≤ q.
• If q < 5
6pp′, we set M = pp′
q hj−1(n); we have ω(M ) = j and ℓ(M ) = p + p′ − q + ℓ(hj−1(n)) ≤ ℓ(hj−1(n)) so that, from (1.12), ℓ(M ) ≤ n holds and (1.9) yields
(3.7) hj(n)≥ M > 6 5hj−1(n) as required. • If q ≥ 5 6pp
By Lemma 2.1, we get p′′ =⋆(q− p) ≥ 107 (q− p), which implies q≤ 10 7 p ′′+ p = pp′′ 7 10 p + 7 p′′ .
But p ≥ 3, p′ ≥ 7, p′′ ≥ p + p′ ≥ 10, thus p′′ ≥ 11, and 10
p + p7′′ ≤ 10 3 + 117 < 356 , yielding q < 56pp′′. By setting M = pp′′ q hj−1(n), as in (3.7), we get hj(n)≥ M > 65hj−1(n). Case 2 : p = 3, p′= 5. • If P+(h
j−1(n)) ≤ 13, which implies n ≤ πid(13) = 41, examining the table of Fig. 1 shows that, for n≤ 41, we have hj(n)≥ 65hj−1(n) with equality if and only if hj−1(n) = 5, 35, 385 or 5005.
• If P+(h
j−1(n)) ≥ 17, and 11 does not divide P+(hj−1(n)), then we apply case 1 with p = 3, p′ = 11, while, if 11 divides P+(hj−1(n)), hj(n)≥ 3· 5 11 hj−1(n) > 6 5hj−1(n) holds.
4
Bounding
h
j(n)
Proposition 4.1. Let j ≥ 1 and n ≥ σj (where σj has been introduced in (1.7)) be two integers; we define r≥ 0 by
(4.1) σj+r− σr≤ n < σj+r+1− σr+1 and n′ by (4.2) 0≤ n′ = n− (σj+r− σr) < pj+r+1− pr+1. Then we have (4.3) hj(n)≤ pr+1pr+2. . . pr+j pj+r+1 pj+r+1− n′ = Nj+r+1 Nr(pj+r+1− n′)· Proof. From its definition (1.9), hj(n) is a product of j primes. Let us denote by A1, A2, . . . , Au (with 0≤ u ≤ j) its prime factors exceeding pj+r+1 and by B1, B2, . . . Br+1+u the primes≤ pj+r+1 and not dividing hj(n); we have
(4.4) hj(n) =
Nj+r+1A1A2. . . Au B1B2. . . Br+1+u
Further, let us introduce ν = ℓ(hj(n)); by (1.12) and (4.1), we have
(4.6) ν ≤ n < σj+r+1− σr+1
and it follows from Proposition 3.2, (3.3), that (4.7) ν = ℓ(hj(n))≥ σj+r− σr. Moreover, (4.4) implies (4.8) ν = ℓ(hj(n)) = σj+r+1− σr+ u X i=1 (Ai− Br+1+i)− r X i=1 (Bi− pi)− Br+1. Now, we consider the optimization problem (where ν, r, u, A1, A2, . . . , Au are fixed)
(4.9) M = max
Z∈D
A1A2. . . Au f (Z)
where D is a subset of Nr+1+u, Z = (Z1, Z2, . . . , Zr+1+u), f (Z) = Z1Z2. . . Zr+1+u
and the setD is defined by
(4.10) Zi ≥ pi, (1≤ i ≤ r + 1), (4.11) Zi< Zr+1, (1≤ i ≤ r), (4.12) Zr+1< Zr+1+i≤ Ai, (1≤ i ≤ u) and (4.13) U (Z)− R(Z) − Zr+1+ σj+r+1− σr = ν with (4.14) U (Z) = u X i=1 (Ai− Zr+1+i), R(Z) = r X i=1 (Zi− pi). Note that, from (4.5) and (4.8), B∈ D so that (4.4) implies (4.15) hj(n) Nj+r+1 = A1A2. . . Au f (B) ≤ M = maxZ∈D A1A2. . . Au f (Z) · If Z ∈ D, from (4.10), (4.11) and (4.12), it follows that
so that f (Z) does not vanish onD and D is finite. Therefore, the maximum M defined by (4.9) is finite; let C be a point in D where the maximum M is attained. We shall prove that
(4.16) U (C) = R(C) = 0.
For that, first we claim that one of the two numbers U (C), R(C) vanishes. Indeed, assume that U (C)6= 0 and R(C) 6= 0. From (4.10), we have R(C) > 0; thus there exists i0, 1≤ i0 ≤ r, such that
(4.17) Ci0 ≥ pi0+ 1 > pi0.
Similarly, from (4.12), we have U (C) > 0, and there exists i1, 1 ≤ i1 ≤ u such that
(4.18) Ai1 > Ai1 − 1 ≥ Cr+1+i1.
Let us define C′∈ Nr+1+uby
Ci′0 = Ci0 − 1, Cr+1+i′ 1 = Cr+1+i1 + 1, Ci′ = Ci for i6= i0, r + 1 + i1. To prove that C′ ∈ D, we have to check that Ci′0 ≥ pi0 (which follows from
(4.17)), that C′
i0 < Cr+1 (which follows from Ci′0 = Ci0 − 1 < Cr+1− 1),
that Cr+1 < Cr+1+i′ 1 (since Cr+1 < Cr+1+i1 and Cr+1+i1 = Cr+1+i′ 1 − 1),
that Cr+1+i′ 1 ≤ Ai1 (which follows from (4.18)) and that U (C′)− R(C′) =
U (C)−R(C) (which follows from U(C′) = U (C)−1 and R(C′) = R(C)−1). Further, we have f (C′) f (C) = Ci′0Cr+1+i′ 1 Ci0Cr+1+i1 = (Ci0− 1)(Cr+1+i1 + 1) Ci0Cr+1+i1 = 1−(Cr+1+i1− Ci0+ 1) Ci0Cr+1+i1 < 1 (4.19)
since, from the definition ofD (cf. (4.11) and (4.12)), Ci0 < Cr+1< Cr+1+i1
holds. But (4.19) contradicts the fact that the maximum in (4.9) is attained at C.
Let us show now that it is impossible to have simultaneously U (C) > 0 and R(C) = 0; indeed, let us assume that U (C)≥ 1 and R(C) = 0 (which implies r = 0 or Ci = pi for 1 ≤ i ≤ r). We define i1 as in (4.18). Since C∈ D, we get from (4.13)
Cr+1 = σj+r+1− σr− ν + U(C) = (σj+r+1− σr+1− ν) + pr+1+ U (C) which, by (4.6) and U (C)≥ 1, yield
We define C′∈ Nr+1+uby
Cr+1′ = Cr+1− 1, Cr+1+i′ 1 = Cr+1+i1 + 1, Ci′ = Ci for i6= r + 1, r + 1 + i1. To prove that C′ ∈ D, we have to check that Ci′ ≥ pi for 1 ≤ i ≤ r + 1 (which follows from Ci′ = Ci = pi if i ≤ r and from (4.20) if i = r + 1), that Ci′ < Cr+1′ for 1 ≤ i ≤ r (which follows from C′
i = Ci = pi ≤ pr and from Cr+1′ = Cr+1− 1 ≥ pr+1, via(4.20)), that Cr+1′ < Cr+1+i′ for 1≤ i ≤ u (which follows from Cr+1′ < Cr+1and Cr+1+i′ ≥ Cr+1+i), that Cr+1+i′ 1 ≤ Ai1
(which follows from (4.18)) and that U (C′)− C′
r+1 = U (C)− Cr+1) (which is easy). As in (4.19), we have f (C′) < f (C), contradicting the fact that the maximum in (4.9) is attained at C.
To prove (4.16), it remains to show that we cannot have R(C) > 0 and U (C) = 0. Let us suppose that R(C) ≥ 1 and U(C) = 0, which implies u = 0 or, for 1≤ i ≤ u,
(4.21) Cr+1+i= Ai ≥ pj+r+2≥ pj+r+1+ 2, with the help of (4.5). From (4.13) and (4.7), this time we get
Cr+1= σj+r+1− σr− ν − R(C) ≤ σj+r+1− σr− (σj+r− σr)− R(C) = pj+r+1− R(C) ≤ pj+r+1− 1. (4.22)
Here we choose i0 as in (4.17) and set
Ci′0 = Ci0− 1, Cr+1′ = Cr+1+ 1, Ci′ = Ci for i6= i0, r + 1.
To prove that C′ ∈ D, we have to check that C′
i0 ≥ pi0 (which follows
from (4.17)), that Cr+1′ ≥ pr+1 (which follows from Cr+1′ = Cr+1 + 1 and Cr+1 ≥ pr+1), that, for 1≤ i ≤ r, Ci′ < Cr+1′ (which follows from Ci′ ≤ Ci and Cr+1′ > Cr+1), that, for 1≤ i ≤ u, Cr+1′ < Cr+1+i1 = Ai1 (which follows
from (4.21) and (4.22)) and that R(C′)+Cr+1′ = R(C)+Cr+1 (which is easy). As precedingly in (4.19), we observe that f (C′) < f (C), contradicting the fact that the minimum is attained at C.
In conclusion, we have proved (4.16) so that Ci = pi for 1≤ i ≤ r and Cr+1+i = Ai for 1≤ i ≤ u. Moreover, (4.6) yields ν ≤ n, and, from (4.13) and (4.2), we get
Cr+1= σj+r+1− σr− ν ≥ σj+r+1− σr− n = pj+r+1− n′. Therefore, the maximum M in (4.9) satisfies
M = A1A2. . . Au
p1p2. . . prCr+1A1A2. . . Au ≤
1
Proposition 4.2. With the notation of Proposition 4.1 , we have (4.23) hj(n)≥ Nj+r+1 Nr(pj+r+1− n′)⋆ = Nj+r+1 qNr
where q = (pj+r+1− n′)⋆ is the smallest prime satisfying q ≥ pj+r+1− n′. Proof. From (4.2), we have pr+1 < pj+r+1 − n′ ≤ pj+r+1 which implies pr+1 ≤ q ≤ pj+r+1 so that M = Nj+r+1/(qNr) is an integer with exactly j prime factors. Further, by (4.2), we have
ℓ(M ) = σj+r+1− σr− q ≤ σj+r+1− σr− (pj+r+1− n′) = n and, by (1.9), hj(n)≥ M holds.
Corollary 4.1. We keep the notation of Proposition 4.1; if q = pj+r+1− n′ is prime then
(4.24) hj(n) = hj(σj+r+1− σr− q) =
Nj+r+1 qNr · Proof. Corollary 4.1 follows from Propositions 4.1 and 4.2.
5
A parity phenomenon
Proposition 5.1. Let k ≥ 2 be an integer and a be an even number satisfying 4≤ a < pk+1 and hk defined by (1.9). We have
(5.1) hk(σk+1− a) = hk(σk+1− a − 1). Proof. Since n 7→ hk(n) in non-decreasing, we have (5.2) hk(σk+1− a) ≥ hk(σk+1− a − 1).
Let us set n = σk+1 − a and note that n satisfies σk < n < σk+1 so that, from (1.8), k = k(n) = k(n− 1). Let M be a positive squarefree integer such that ℓ(M )≤ n and ω(M) = k. Such a number M is even; if not, we would have ℓ(M ) ≥ 3 + 5 + . . . + pk+1 = σk+1 − 2 in contradiction with ℓ(M )≤ n = σk+1− a ≤ σk+1− 4. Therefore, ℓ(M) is the sum of 2 and k − 1 odd numbers, so that ℓ(M )≡ σk ≡ σk+1+ 1≡ σk+1− a − 1 (mod 2). So, ℓ(M ) cannot be equal to σk+1− a and ℓ(M) ≤ σk+1− a − 1 holds. Thus, from (1.9), we get hk(σk+1− a) ≤ hk(σk+1− a − 1), which, with (5.2), proves (5.1).
Proposition 5.2. Let k be an integer, k ≥ 2, and q a prime number satis-fying 3 ≤ q ≤ pk. By setting m = σk+1− q − 1, we have
(5.3) hk−1(m) = hk−1(m− 1) = hk−1(σk+1− q − 2) = Nk+1
Proof. From the table of Figure 1, we have h1(6) = h1(5) = 5, h2(11) = h2(10) = 21 and h2(13) = h2(12) = 35 so that the proposition is true for k = 2, q = 3 and for k = 3 and q = 3 or 5. So, from now on, we assume k≥ 4. Corollary 4.1 with j = k − 1, r = 1 implies hk−1(m− 1) = Nk+1/(2q) and, since n7→ hk−1(n) is non-decreasing, it follows that
(5.4) hk−1(m)≥ hk−1(m− 1) = Nk+1
2q ·
Let M be a positive squarefree integer satisfying ℓ(M ) ≤ m and ω(M) = k− 1. In view of (5.4) and (1.9), to prove that hk−1(m) = Nk+1/(2q), it suffices to show that
(5.5) M ≤ Nk+1
2q ·
If M is odd, ℓ(M ) is the sum of k− 1 odd numbers, which implies ℓ(M )≡ σk≡ σk+1− q = m + 1 (mod 2).
So, ℓ(M ) cannot be equal to m; since, by (1.9), ℓ(M )≤ m holds, we should have ℓ(M )≤ m − 1; therefore, from (1.9), we get
(5.6) M ≤ hk−1(m− 1) =
Nk+1 2q ·
If M is even, we have ω(M/2) = k− 2 and ℓ(M/2) ≤ m − 2, so that
(5.7) M ≤ 2hk−2(m− 2).
• If q ≥ 11, since we have assumed k ≥ 4, i.e. pk+1≥ 11, we have σk− 5 ≤ m − 2 = σk+1− q − 3 ≤ σk+1− 14 < σk+1− 10. By Proposition 4.1 with j = k−2, r = 2, n = m−2, n′= n−(σ k−σ2) = pk+1− q + 2, we get hk−2(m− 2) ≤ Nk+1 6(q− 2) which, by (5.7), gives (5.8) M ≤ Nk+1 3(q− 2) = Nk+1 3q q q− 2 ≤ Nk+1 3q 11 9 < Nk+1 2q · • If q ∈ {3, 5, 7}, since k ≥ 4 and pk+2 ≥ p6 = 13, we have
and Proposition 4.1 with j = k− 2, r = 3, n = m − 2 and n′ = n− (σk+1− 10) = 7 − q yields
(5.9) hk−2(m− 2) ≤
Nk+2 30(pk+2+ q− 7)· Using q≤ 7, k ≥ 4 and pk+2≥ p6= 13 gives
Nk+2 30(pk+2+ q− 7) = Nk+1 30 pk+2 pk+2+ q− 7 ≤ Nk+1 30 13 q + 6 = 13Nk+1 30q q q + 6 ≤ 13Nk+1 30q 7 13 < Nk+1 4q which, together with (5.7) and (5.9), proves
(5.10) M < Nk+1
2q ·
Inequalities (5.6), (5.8) and (5.10) prove that (5.5) holds, which, with (5.4), completes the proof of (5.3).
Proposition 5.3. Let k be a positive integer and m = σk+1− 1; we have (5.11) hk(m) = hk(m− 1) = hk(σk+1− 2) =
Nk+1 2 ·
Proof. It is the same proof than for Proposition 5.2. By Proposition 3.2 with j = k and r = 1, we have
hk(m)≥ hk(m− 1) = Nk+1
2 ·
6
The increasingness of
h
j(n) on j
Theorem 6.1. Let n ≥ 2 be an integer and k = k(n) be defined by (1.8); for j satisfying 1 ≤ j ≤ k, we have
(6.1) hj−1(n)≤
5 6hj(n)
and (6.1) is an equality if and only if j = k(n) ≥ 2 and n = σj+1− 4 or n = σj+1− 5.
Proof. If j = 1, it follows from (1.10) and (1.11) that h0(n) = 1, h1(n) = ⋆(n)≥ 2 and h
0(n)/h1(n) ≤ 1/2 < 5/6, which proves (6.1). So, from now on, we assume j≥ 2.
The sequence (σj+r− σr)r≥0 is increasing and goes to infinity. So, we may define rj ≥ 0 and n′j by
(6.2) σj+rj− σrj ≤ n < σj+rj+1− σrj+1
and
(6.3) n′j= n− (σj+rj− σrj).
We shall consider four cases : rj ≤ j − 4, rj ≥ j + 3, j − 3 ≤ rj ≤ j + 2 and j≥ 25, j − 3 ≤ rj ≤ j + 2 and j ≤ 24.
First case : rj ≤ j − 4
From (2.5) and our hypothesis j≥ rj+ 4, we deduce (6.4) prj+1+ prj+2≤ p2rj+3< pj+rj < pj+rj+1
and
(6.5) prj+2+ prj+3≤ p2rj+5< pj+rj+2.
Let us set From (6.2), we get
(6.6) 0≤ n′j = n− (σj+rj − σrj) < pj+rj+1− prj+1
and applying Proposition 4.2 yield
(6.7) hj(n)≥
Nj+rj+1
qNrj
= Nj+rj+1 Nrj(pj+rj+1− n′j)⋆·
In view of bounding hj−1(n), we have to determine rj−1 such that (6.8) σj−1+rj−1 − σrj−1 ≤ n < σj+rj−1− σrj−1+1.
Sub case one, rj−1= rj+ 1 Let us asume that
(6.9) σj+rj − σrj ≤ n < σj+rj+1− σrj+2.
i.e. from (6.3),
(6.10) 0≤ n′j = n− (σj+rj− σrj) < pj+rj+1− prj+1− prj+2.
Note that, from (6.4), the right hand side of (6.10) is positive. Then, we have rj−1= rj+ 1 since, from (6.9),
σ(j−1)+(rj+1)− σrj+1 = σj+rj− σrj+1 < σj+rj− σrj ≤ n
< σj+rj+1− σrj+2 = σ(j−1)+(rj+1)+1− σ(rj+1)+1
holds. Via (6.3), this implies that
n′j−1== ndef − (σj−1+rj−1− σrj−1) = n− σj+rj+ σrj+1= n′j + prj+1.
Applying Proposition 4.1 and noting that j− 1 + rj−1= j + rj yield hj−1(n)≤ Nj−1+rj−1+1 Nrj−1(pj−1+rj−1+1− n′j−1) = Nj+rj+1 Nrj+1(pj+rj+1− n′j− prj+1) · By using (6.7), we get (6.11) hj−1(n) hj(n) ≤ (pj+rj+1− n′j)⋆ prj+1(pj+rj+1− n′j− prj+1) ·
From (6.6), we have pj+rj+1− n′j > prj+1 ≥ p1 = 2, so that we may apply
Lemma 2.1 which, with the help of (6.11) and (6.10), yields hj−1(n) hj(n) ≤ 11 8prj+1 1 + prj+1 pj+rj+1− n′j− prj+1 ! < 11 8 1 prj+1 + 1 prj+2 . If rj ≥ 1, 118 1 prj+1 +prj+21 ≤ 118 13 +15 < 56, which proves (6.1).
It remains to consider the case rj = 0, which, from (6.9) and (1.8), implies σj ≤ n < σj+1 and k(n) = j.
By Lemma 2.1, (a)⋆ ≤ 11
8 a holds, and, for a≥ 12, 11 16 a a− 2 ≤ 11 16 12 10 < 5 6, which, via (6.13), proves (6.1).
Since, from (6.12), a > 5, it remains to study the cases 6 ≤ a ≤ 11. If a = 7, 9, 10, 11, it is easy to check that (a)
⋆ 2(a− 2) <
5 6·
If a = 6 or a = 8, by Proposition 5.1, (6.12) and (4.23) we have hj(n) = hj(σj+1− a) = hj(σj+1− a − 1) = hj(n− 1) ≥
Nj+1 (a + 1)⋆ while, by Proposition 5.2, since a− 1 is prime, we get
hj−1(n) = hj−1(n− 1) = Nj+1 2(a− 1) yielding hj−1(n) hj(n) = hj−1(n− 1) hj(n− 1) ≤ (a + 1)⋆ 2(a− 1) = ( 7/10 if a = 6 11/14 if a = 8 and, in both cases, hj−1(n)
hj(n) < 5
6 holds, which proves (6.1). Sub case two, rj−1 = rj+ 2
Now, we asume that (6.2) holds but not (6.9); thus we have (6.14) σj+rj+1− σrj+2 ≤ n < σj+rj+1− σrj+1
and, from (6.5),
(6.15) pj+rj+1− prj+1− prj+2≤ n′j < pj+rj+1− prj+1.
Here, we get rj−1= rj+ 2, since we have
σj−1+rj+2− σrj+2 ≤ n < σj+rj+1− σrj+1 < σj+rj+2− σrj+3
by observing that, from (6.5),
σj+rj+2− σrj+3− (σj+rj+1− σrj+1) = pj+rj+2− prj+2− prj+3> 0
holds. Now, we have
0≤ n′j−1 = n− (σj−1+rj−1− σrj−1) = n− σj+rj+1+ σrj+2
= n′j+ σj+rj− σrj − σj+rj+1+ σrj+2 from (6.6)
= n′j− pj+rj+1+ prj+1+ prj+2< prj+2 from (6.15)
and applying Proposition 4.1 gives hj−1(n)≤ Nj−1+rj−1+1 Nrj−1(pj−1+rj−1+1− n′j−1) = Nj+rj+2 Nrj+2(pj+rj+2− n′j−1) while Proposition 4.2 yields
hj(n)≥
Nj+rj+1
Nrj(pj+rj+1− n′j)⋆·
We set a = pj+rj+1 − n′j and ∆ = pj+rj+2 − prj+1− prj+2 so that (6.16)
allows to write pj+rj+2− n′j−1= ∆ + a, and we have
(6.17) hj−1(n) hj(n) ≤ pj+rj+2 prj+1prj+2 (a)⋆ ∆ + a· (6.16) can be rewritten as (6.18) 2 = p1≤ prj+1 < a = pj+rj+1− n′j ≤ prj+1+ prj+2.
Lemma 2.1 implies (a)⋆ ≤ 11a/8 and, by (6.4), ∆ > 0 holds, so that the homographic function t7→ t/(∆ + t) is increasing. From (6.18), we thus have
a ∆ + a ≤ prj+1+ prj+2 ∆ + prj+1+ prj+2 = prj+1+ prj+2 pj+rj+2 · Therefore, we get hj−1(n) hj(n) ≤ 11 8 prj+1+ prj+2 prj+1prj+2 = 11 8 1 prj+1 + 1 prj+2
which is smaller than 5
6 if rj ≥ 1.
It remains to consider the case rj = 0, rj−1 = 2. Formula (6.17) becomes
(6.19) hj−1(n) hj(n) ≤ pj+2 6 (a)⋆ pj+2− 5 + a while (6.18) via (6.3) becomes
(6.20) 2 < a = pj+1− n′j = σj+1− n ≤ 5.
Since j ≥ 2 holds, note that (6.20) implies σj ≤ σj+1− 5 ≤ n < σj+1− 2, which shows from (1.8) that k(n) = j.
• If a = 5, since n = σj+1−a = σj+1−5, by Corollary 4.1 with r = rj = 0 and q = 5, we get hj(n) = Nj+1/5, while Proposition 3.2 gives
• If a = 4, by Proposition 5.1, we get hj(σj+1− 4) = hj(σj+1− 5) and, by Proposition 5.2, hj−1(σj+1− 4) = hj−1(σj+1− 5) so that hj−1(σj+1− 4) hj(σj+1− 4) = hj−1(σj+1− 5) hj(σj+1− 5) = 5 6· • If a = 3, Formula (6.19) becomes hj−1(n) hj(n) ≤ pj+2 2(pj+2− 2) ≤ 7 10 < 5 6 since pj+2≥ p4= 7. Second case : rj ≥ j + 3
From (6.2), we deduce n ≥ σj+rj − σrj+1 = σ(j−1)+(rj+1) − σ(rj+1) and
Proposition 3.2, (3.3), implies
(6.21) ℓ(hj−1(n))≥ σj+rj− σrj+1.
Let us now show that
(6.22) q == Pdef +(hj−1(n))≥ pj+rj.
Indeed, if q≤ pj+rj−1holds, since hj−1(n) has j−1 prime factors, we should
have
ℓ(hj−1(n)) ≤ pj+rj−1+ pj+rj−2+ . . . + prj+1
< pj+rj+ pj+rj−1+ . . . + prj+2 = σj+rj− σrj+1
which would contradict (6.21).
Further, among the j + 1 primes p2 = 3, p3, . . . , pj+2, there are certainly two primes p and p′ not dividing hj−1(n) and satisfying 3≤ p < p′ ≤ pj+2. By Lemma 2.4, (2.5), and (6.22), we get
(6.23) p + p′ ≤ pj+1+ pj+2≤ p2j+3 ≤ pj+rj ≤ q = P
+(h
j−1(n)) and, applying Proposition 3.4 proves hj−1(n)
hj(n) < 5
6·
Third case : j− 3 ≤ rj ≤ j + 2 and j ≥ 25
The proof is the same than for the second case; only, in (6.23), instead of (2.5), we use (2.6) with b = 7, i = j + 2≥ 27 :
p + p′ ≤ pj+1+ pj+2< p2j−3 ≤ pj+rj ≤ q = P
+(h
Fourth case : j− 3 ≤ rj ≤ j + 2 and j ≤ 24
Here, we have rj ≤ j + 2 ≤ 26 and, from (6.2), we get n < σj+rj+1≤ σ51= 5350.
So, for k≤ 50, σk ≤ n < σk+1 and 1≤ j ≤ k, we have computed hj(n) with the algorithm described in Section 1.4 and we have checked that, for j ≥ 2,
hj−1(n) hj(n) ≤
5 6
always holds, with equality if and only if j = k(n) and n = σj+1− 4 or n = σj+1− 5.
Corollary 6.1. For all non-negative integer n ≥ 2, we have
(6.24) h(n) = hk(n)
where k = k(n) is defined by (1.8). Proof. From (1.1) and (1.9) we have
h(n) = max
0≤j≤k(n)hj(n)
and Theorem 6.1 yields h0(n) < h1(n) < . . . < hk(n)(n).
7
Computation of
π
f(x)
Let f be an arithmetic function, i.e a function defined on positive integers. The simplest way to compute πf(x) defined in (1.5) is to generate the primes up to x by Eratosthenes’s sieve, which is too expensive for large values of x. Definition 7.1. An arithmetic function f is said to be completely multi-plicative if f (ab) = f (a)f (b) for all a and b. If f 6= 0, this implies f(1) = 1. Following ideas of the german astronomer Meissel, Lagarias, Miller and Odlyzko gave in [8] an algorithm that computes π(x) with a cost O x
2/3 log x
! . In this work they also remark that their algorithm allows to compute πf(x) for every completely multiplicative arithmetic function f .
This method has been improved in [4] to compute π(x) with a cost
O x
2/3 log2x
!
, provided that all the arithmetic operations on integers are of constant cost O(1), not depending on the size of the operands. We show here that this improved algorithm may be used to compute πf(x) whith a cost which is still O x
2/3 log2x
!
Proposition 7.1. Let f be a completely multiplicative arithmetic function with integer values. Let F be the summatory function of f ,
(7.1) F (x) =X
n≤x f (n).
We suppose that all the ordinary arithmetic operations about integers are of constant cost O(1), and that
1. Each value f (n) may be computed in time O(1), not depending of the size of n.
2. There is an algorithm computing
(7.2) S0(y, x) = X 1≤n≤y µ(n)f (n)F x n in time Ox2/3/ log2x.
Then, there is an algorithm computing πf(x) = X p≤x p prime f (p) in time O x 2/3 log2x ! . When F (u) can be computed in O(1) time, the second hypothesis is sat-isfied.
Remarks :
1. The second hypothesis may seem strange. Let us give a few words of explanation.
• Our computation of πf(x) begins by choosing y = O x1/3+ε. Then we compute S0 = S0(y, x) (this is the contribution of ordi-nary leaves defined in lemma 5.2, equation (9) in [4] and in lemma 7.2, equation (7.14) in this article). Function F does not appear elsewhere in the algorithm. S0 being computed, the total cost of the other computations is Ox2/3/ log2x. Condition (2) ensures that our algorithm computes πf(x) in time O
x2/3/ log2x. • In many cases, F (u) can be computed in time O(1), then the sum
(7.2) can be computed in time O(y), by precomputing the Möbius function, so that the second hypothesis is satisfied.
2. In Proposition 7.1 we restrict ourseves to the case of integer valued functions. The case of real valued functions is more delicate because of truncation errors. In [1], Bach and al. have elaborated an algorithm to compute πf(x) where f (n) = 1/n, and
Algorithm for πf(x)
We will describe very briefly our algorithm to compute πf(x), using notations and formulas which, when replacing f by 1, reduce to the correponding ones contained in [4].
For b∈ N, let us define Φ(x, b) as the sum of the f(n), for the n′s∈ [1, x] that subsist after sieving this interval by all primes p1, p2, . . . , pb,
(7.3) Φ(x, 0) = X 1≤n≤x f (n) = F (x) and, for b≥ 1, Φ(x, b) = X 1≤n≤x P −(n)>pb f (n)
For k≥ 1 and b ≥ 1, let us set
(7.4) Pk(x, b) =
X
1≤n≤x Ω(n)=k, P −(n)>pb
f (n).
so that, from (7.3) and (7.4) we get, for x≥ 1,
(7.5) Φ(x, b) = 1 + P1(x, b) + P2(x, b) +· · · From now on, we choose y ∈ R
(7.6) x1/3 ≤ y ≤√x and set a = π(y).
We will precise later the best choice for y, which is closed to x1/3. Since y ≥ x1/3 equation (7.4) yields P
k(x, a) = 0 for k ≥ 3 and (7.5) becomes Φ(x, a) = 1 + P1(x, a) + P2(x, a). Since P1(x, a) = X pa<p≤x f (p) = X y<p≤x f (p) = πf(x)− πf(y), (7.7) πf(x) = Φ(x, a) + πf(y)− 1 − P2(x, a).
Replacing f by 1 (and πf by π), formula (7.7) is formula (4) in [4].
7.1 Initialization of the computation: the 2 basis tables
7.2 Computation of P2(x, a)
Definition (7.4) and the complete multiplicativity of f give P2(x, a) = X y<p≤q≤x pq≤x f (pq) = X y<p≤q≤x pq≤x f (p)f (q)
where p and q are primes. The p′s figuring in this sum satisfy p ≤ x q ≤ x y and we get P2(x, a) = X y<p≤x/y f (p) X p≤q≤x/p f (q).
We remark that, for p >√x, the sum on q vanishes. Since, by (7.6),√x≤ x y, we have P2(x, a) = X y<p≤√x f (p) X p≤q≤x/p f (q) = X y<p≤√x f (p) πf x p − πf(p− 1) or (7.8) P2(x, a) = X y<p≤√x f (p)πf x p − X y<p≤√x f (p)πf(p− 1).
In the above formula, the values of p are bounded above by√x which is larger than y. Thus we cannot find these primes p, nor the values πf(p− 1) in the precomputed tables (cf. §7.1), and we generate them using a sieve of [1,√x], which we call the auxilliary sieve. The values of x/p lie in the interval [1, x/y]. So we will get the values πf(x/p) by an other sieve, the main sieve. Let us note that the respective sizes of the sieve intervals, √x and x/y, are too large to allow a sieve in one pass. Thus the two sieves will be done by blocks of size y that must be synchronized.
– substract f (p) from T . Thus the new value of T is πf(p− 1). – If x/p > B, while x/p > B we replace the block [A, B] by the next
block [A + y, B + y] and we sieve it. When x/p∈ [A, B] we get πf(x/p) in the main sieve table and we add f (p)πf(x/p)− f(p)T to P2.
– Using the auxilliary sieve, replace p by its predecessor.
The final value of the variable P2 is P2(x, a). The first step is negligible in cost, compared to the second. Thus the computation of P2(x, a) is of total cost O x y log log y . 7.3 Computation of Φ(x, a)
The following lemma is proved as lemma 5.1 in [4]. Lemma 7.1. For every u ≥ 0, and for b ≥ 1,
(7.9) Φ(u, 0) = F (u) (7.10) Φ(u, b) = Φ(u, b− 1) − f(pb)Φ u pb , b− 1
This relation gives an obvious method for computing Φ(x, a). Starting from the tree with the only node Φ(x, a), and applying repeatedly (7.10) we get a tree whose all nodes, except the root node, are labelled by a formula of the form
(7.11) µ(n)f (n)Φ x
n, b
where b ≤ a − 1 and n = 1 or n is a squarefree integer with prime factors q∈ {pb+1, . . . , pa}.
If we repeat this expansions until all the leaves of the resulting tree are labelled by formulas µ(n)f (n)Φ xn, 0, using (7.9) we get the formula : (7.12) Φ(x, a) = X 1≤n≤x P +(n)≤y µ(n)f (n)Φ x n, 0 = X 1≤n≤x P +(n)≤y µ(n)f (n)F x n
which, when f = 1 is formula Φ(x, a) = X
1≤n≤x P +(n)≤y µ(n)j x n k (cf. [4, p. 237]).
The number of terms in (7.12) is much too large. In order to get a sum with fewer terms we replace the trivial rule
which leads to (7.12) by the new rule
Rule 2 : Expand node(7.11) only if b > 0 and n ≤ y. Expanding the computation tree whith rule 2 instead of rule 1 we get Lemma 7.2. We have
(7.13) Φ(x, b) = S0+ S,
where S0 is the contribution of ordinary leaves (7.14) S0 = S0(y, x) = X 1≤n≤y µ(n)f (n)Φ x n, 0 = X 1≤n≤y µ(n)f (n)F x n
and S, the contribution of special leaves, is
(7.15) S = X n P −(n)≤y<n µ(n)f (n)Φ x n, π(P −(n))− 1·
This lemma corresponds to lemma (5.2) in [4]. 7.3.1 Computation of S0
In the general case, the computation of S0is done with a cost O x2/3/ log2x thanks to the condition 2 in proposition 7.1.
In the case we will consider later in this work, the computation of πid(x), f (n) = n, thus F (u) = [u][u + 1]
2 is computed in O(1) time and the compu-tation of S0(x, y) is of cost O(y) = o
x2/3/ log2x. 7.3.2 Computation of S
In the sum (7.15), let us set n = mp with p = P−(n). Grouping together all the n′s according to the value of p, we get
(7.16) S =−X p≤y f (p) X P −(m)>p m≤y<mp µ(m)f (m)Φ x mp, π(p)− 1 ·
7.3.3 How to compute S in Ox2/3+ε
In this section, we explain a first method to get πf(x), rather simple to implement, and whose running time is O x2/3+ε. We take y = x1/3. Since mp > y all the values u = x/mp appearing in (7.16) are less than x2/3. We sieve the interval
1, x
y
successively by all primes p ≤ y. After the sieve by p, from the definition (7.3) of Φ, for all the m’s such that m≤ y < mp, we get in the sieve table the value Φ
x
mp, π(p)− 1
, and we add to S the value f (p)µ(m)f (m)Φ x mp, π(p)− 1 ·
But, if we proceed in the naive way, after sieving by each p, we will update the sieve table, putting in the case of index u the sum of f (n) for the n’s, n≤ u that are still in the table. This is excluded because, for each p this would cost O(x/y) operations, and the total cost of these updatings would be≫ π(y)(x/y) = x/ log x. As explained in [8] (the 7 last-lines p. 545 and the first half of p. 546) we use an auxiliary data structure such that, for a price of O(log x) time in place of O(1) for each access, we don’t need to update the sieve table after each sieve. To be a little more precise let us say that this structure is a labelled binary tree. There is a leave for each index i of the table sieve, this leave is labelled by the value f (i), and each interior node is labelled by the sum of labels of its two sons. Proceeding in this way the cost of the sieve is O
x
y log x log log x
, while the cost of retrieving the values f (p)µ(m)f (m)Φ
x
mp, π(p)− 1
is O(π(y)y log x). Both costs are Ox2/3+ε with our choice y = x1/3.
7.3.4 Faster computation of S
In this section, we explain how to carry out the computation of πf(x) in
O x
2/3 log2x
!
. We take y = x1/3(log x)3log log x. To speed up the computa-tion of S we particomputa-tion (7.16) in 3 subsums S = S1+ S2+ S3,
S3 = − X p≤x41 f (p) X P −(m)>p m≤y<mp µ(m)f (m)Φ x mp, π(p)− 1
We will show that S1 is quickly computed in O(y) time. S3 will be com-puted by sieve, as explained in §7.3.3, but faster because the number of values for p is reduced from π(y) to π(x1/4). The main part of the computation will be the computation of S2.
As in [4], we first observe that the m′s involved in S1and S2are all prime and therefore : S1 = X x13<p≤y f (p) X p<q≤y f (q)Φ x pq, π(p)− 1 (7.17) S2 = X x14<p≤x31 f (p) X p<q≤y f (q)Φ x pq, π(p)− 1 (7.18)
ComputingS1 As in [4] we remark that, in (7.17), we have x pq < x
1/3 < p.
Thus, all the values Φ x
pq, π(p)− 1
are equal to 1. Therefore S1= X x13<p≤y f (p) X p<q≤y f (q) = X x31<p≤y f (p) (πf(y)− πf(p)) .
This value is computed in O(y) additions, using the precomputed table of the values πf(u) for 1≤ u ≤ y.
ComputingS3 For each p≤ x1/4we precompute the list of all the square-free m≤ y whose least factor is p.
We sieve the interval h1, xyi successively by all the primes up to x1/4. As soon as we have sieved by p, using the precomputded lists of squarefree whose least prime factor is a prime q > p we sum the
f (p) X P −(m)>p m≤y<mp µ(m)f (m)Φ x mp, π(p)− 1
Thus the cost of sieving is O
x
ylog x log log x
. The number of values of p is π x1/4and the number of values of m is less than y, thus the cost of re-trieving the values f (p)µ(m)f (m)Φ
x
mp, π(p)− 1
is Oπ x1/4× log x × y Thus computing S3 is of cost O
x
ylog x log log x + yx 1/4
Computing S2 We split the sum (7.18) in two parts depending on q > x/p2 or q≤ x/p2. It gives S2 = U + V with U = X x14<p≤x13 f (p) X p<q≤y q>x/p2 f (q)Φ x pq, π(p− 1 and V = X x41<p≤x13 f (p) X p<q≤y q≤x/p2 f (q)Φ x pq, π(p− 1
Computing U With y <√x (cf. (7.6)), the condition q > x/p2 implies p2 > x/q≥ x/y ≥ x1/2. Thus, U = X √ x/y<p≤x1/3 f (p) X p<q≤y q>x/p2 f (q)Φ x pq, π(p− 1
From x/p2 < q we deduce x/pq < p and Φ(x/pq, π(p)− 1) = 1, and we have x/p2≥ p so that U = X √ x/y<p≤x1/3 f (p) X p<q≤y q>x/p2 f (q) = X √ x/y<p≤x1/3 f (p) πf(y)− πf x p2 ·
Since x/p2 < q ≤ y the sum U is calculated in O(y) operations with the table of values of πf(u).
Computing V For each term involved in V we have p≤ x pq < x
And we write V = V1+ V2 with V1 = X x1/4≤p<x1/3 f (p) X p<q≤min(x p2,y) f (q)(1− πf(p− 1)) V2 = X x1/4≤p<x1/3 f (p) X p<q≤min(x p2,y) f (q)πf x pq
Computing V1 can be achieved in O(y) time once we have tabulated πf(u) for u≤ y.
Computing V2. We first split V2 in two parts in order to simplify the condition q≤ min(x/p2, y) : V2 = X x1/4<p≤qx y f (p) X p<q≤y f (q)πf x pq + X qx y<p<x1/3 f (p) X p<q≤x p2 f (q)πf x pq
In the purpose to speed up the computation of the above two sums we now write, V2 = W1+ W2+ W3+ W4+ W5 with W1 = X x1/4<p≤x y2 f (p) X p<q≤y f (q) πf x pq W2 = X x y2<p≤ qx y f (p) X p<q≤qx p f (q) πf x pq W3 = X x y2<p≤ qx y f (p) X qx p<q≤y f (q) πf x pq W4 = X qx y<p≤x1/3 f (p) X p<q≤qxp f (q) πf x pq W5 = X qx y<p≤x1/3 f (p) X qx p<q≤ x p2 f (q) πf x pq
The cost of this computation is the sum of three terms : • The cost of the above sieve on 1,√xis O √x log log x. • The cost of adding the terms of the sum W1, O
x y log2x
.
• The cost of adding the terms of the sum W2, O
x3/4 y1/4log2x
! .
ComputingW3 and W5 For W3, for each p we apply lemma 2.6 with z = x/p and u = y. Thus, for each value of p, the sum on q costs Oπ(px/p), and the total cost of the computation of W3 is
O X x y2<p≤ x y π r x p .
For W5, for each p we apply lemma 2.6 with z = x/p and u = x/p2. Thus, for each value of p, the sum on q costs Oπ(px/p), and the total cost of the computation of W5 is O X qx y<p≤x1/3 π r x p . Thus the costs of computing W3 and W5 add to
O X x y2<p≤x 1/3 π r x p = O X x y2<p≤x 1/3 q x p logx p = O x 2/3 log2x ! ·
ComputingW4 We simply sum over (p, q). There would be no advantage to proceed as for W3 since most of the values πf(x/pq) are distinct. The cost is O X qx y<p≤x1/3 π r x p = O x2/3 log2x ! ·
8
The algorithm to calculate
h(n)
8.1 The function G(pk, m)
The function G(pk,m) has been introduced and studied in [5].
Definition 8.1. Let pk be the k-th prime, for some k ≥ 3 and m an integer satisfying 0 ≤ m ≤ pk+1− 3. We define
(8.1) G(pk, m) = max
Q1Q2. . . Qs q1q2. . . qs
where the maximum is taken over the primes Q1, Q2, . . . , Qs, q1, q2, . . . , qs (s ≥ 0) satisfying (8.2) 3≤ qs < qs−1 < . . . < q1 ≤ pk< pk+1≤ Q1< Q2 < . . . < Qs and (8.3) s X i=1 (Qi− qi)≤ m.
The additive function ℓ (cf. §1.1) can easily be extended to fractions by setting
ℓ(M/N ) = ℓ(M )− ℓ(N)
when M and N are coprime or are both squarefree. Therefore, the inequality (8.3) implies
(8.4) ℓ(G(pk, m))≤ m.
8.1.1 Properties of G(pk, m)
Obviously, G(pk, m) is non-decreasing on m and G(pk, 2m + 1) = G(pk, 2m). The maximum in (8.1) is unique (from the uniqueness of the standard fac-torization into primes). For small m’s, we have
(8.5) 0≤ m < pk+1− pk =⇒ G(pk, m) = 1. From Proposition 8 of [5], we have
(8.6) pk+1
(pk+1− m)⋆ ≤ G(pk
, m)≤ pk+1 pk+1− m
.
8.1.2 Computation of G(pk, m)
In [5, §8], two algorithms are given to calculate G(pk, m).
The first one is a combinatorial algorithm. In its first step, the primes allowed to divide the denominator of G(pk, m) are determined. From (8.2) and (8.3), they are all the primes in the range [(pk+1− m), pk], say P1 < P2 < . . . < PK. Similarly, the primes authorized to divide the numerator are all the primes PK+1 < PK+2 < . . . < PR in [pk+1, pk+ m]. By setting P′ ={P1, P2, . . . , PR}, from the definition (1.14), we get
G(pk, m) =
1 P1P2. . . PK
hK(P1+ P2+ . . . + PK+ m,P′)
and hj(n,P′) can be computed by induction on j in a way similar to that exposed in §1.4. In [5, §8], one can find the details and also some tricks to improve the running time of this combinatorial algorithm which, however, remains rather slow when m is large.
The second algorithm, which is more sophisticated, is based on the fol-lowing remark : if G(pk, m) =
Q1Q2. . . Qs q1q2. . . qs
and m is large, the least prime factor qs of the denominator is close to pk+1− m while all the other primes Q1, . . . , Qs, q1, . . . , qs−1 are close to pk.
More precisely, the following proposition (which is Proposition 10 of [5]) says that if pk+1− m + δ is prime for some small δ and if G(pk+1, δ) is not too small, then the computation of G(pk, m) is reduced to the computation of G(pk+1, m′) for few small values of m′, which can be done by the above combinatorial algorithm.
Proposition 8.1. We want to compute G(pk, m) as defined in (8.1) with pk odd and pk+1 − pk ≤ m ≤ pk+1− 3. We assume that we know some even non-negative integer δ satisfying
(8.7) pk+1− m + δ is prime, (8.8) G(pk+1, δ)≥ 1 + δ pk+1 and (8.9) δ < 2m 9 < 2pk+1 9 · If δ = 0, we know from (8.6) that G(pk, m) =
where bq is defined by
(8.11) q =b pk+1pk+2(pk+1− m + δ)
(pk+1+ δ)(pk+1− 3δ/2) ≤ pk+2− m + 3δ
2 ·
How to compute G(pk, m) ? The combinatorial algorithm should be tried if m is small, but it is quadratic in m and has no chance to terminate if m is larger than, say, 106. We have no guarantee that the conditions of Prop. 8.1 are satisfied. However in all our numerical applications, we have found δ < 1000 in (8.7) (see [5, §9.2]), so that, by (8.10) and (8.11), we have
m− pk+1+ q≤ m − pk+1+ bq ≤ pk+2− pk+1+ 3δ
2
and, in (8.10), G(pk+1,m− pk+1+ q) can be easily calculated by the combi-natorial algorithm.
8.2 Description of the algorithm to compute h(n)
To compute h(n), the first step is to determine pk and σk defined by (1.8). This step is explained in § 8.2.1 and will furnish also pk+1 and n′= n− pk. 8.2.1 Computation of pk and σk
1. Compute x =qLi−1(n), so that Li(x2) = n and x∼√n log n. 2. Using Prop. 7.1, we compute πid(x) in time
Ox2/3/ log2x= On1/3/(log n)−5/3.
3. To get σk, we have to add (if πid(x) < n) or to subtract (if πid(x) > n) to πid(x) the primes between x and pk, calculated by sieving. In practice, this step is very short. But we are able to estimate it only under Riemann’s hypothesis. By lemma 2.5, we have
Li(p2k)− 5 24πp 3/2 k log pk< σk ≤ n < σk+1< Li(p2k+1)+ 5 24πp 3/2 k+1log pk+1 which implies n = Li(p2k) + Op3/2k log pk ∼ Li(p2k)∼ p2k 2 log pk· Therefore, we get log n∼ 2 log pk, pk∼√n log n and (8.12) Li(p2k) = n + On3/4(log n)7/4. Further, since x∼ pk∼√n log n, we have
8.2.2 Computation of h(n)/Nk
By Corollary 6.1, we have h(n) = hk(n). Let us set n′ = n− σk. If n′ = pk+1− 1 or n′ = pk+1− 2, Proposition 5.3 yields h(n) = Nk+1/2. So, we may suppose n′ ≤ σk+1− 3. From the definition (8.1) of function G, we have (8.13) h(n) = hk(n) = NkG(pk, n′)
and we compute G(pk, n′) as explained in §8.1.2. In practice, the compu-tation of G(pk, n) is very fast. However, as explained in §8.1.2, we have no estimation of the running time.
Below, are listed some values of h(n) Nk
= G(pk, n′) together with pk, n′ = n− σk, e = e(n) the largest integer such that h(n− e) = h(n)) and, if the algorithm of Proposition 8.1 is used, δ and Q, the number of primes used in the sum (8.10).
n = 1012, pk= 5477081, n′ = 4935150, e = 0, δ = 18, Q = 1, G(pk, n′) = 29998525822277 2968309525031 = 5477089× 5477093 5477081× 541951 · n = 1035, pk= 2898434150644708999, n′ = 1886081812111845520, e = 16 δ = 134, Q = 5, G(pk, n′) = 2898434150644709023 1012352338532863519·
The values of h(10a) for a≤ 35 and of h = 2b for b≤ 116 can be found on the authors’s web sites [2, 10], together with the Maple or Sage programs computing h(n)/Nk.
9
An open question
Given n and j < k(n), how to compute hj(n)? We have not succeeded in solving this problem when n is too large to use the naive algorithm described in 1.4. The case j = 2 is already not that simple.
A first step is certainly to calculate r = r(n, j) defined by (4.1), which can be done by the method of § 8.2.1. If we are lucky enough that q = pj+r+1−n′ = pj+r+1−(n−σj+r+σr) is prime, then the value hj(n) = Nj+r+1qNr is given by (4.24).
In the general case, by setting n′ = n− (σj+r − σr), one may think that h′j(n) = Nj+r
Nr
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Marc Deléglise, deleglis@math.univ-lyon1.fr. Jean-Louis Nicolas, jlnicola@in2p3.fr.
Université de Lyon, CNRS,
h1 h2 h3 h4 h5 h6 h n = 2 2 2 3 3 3 4 3 3 5 5 6 6 6 5 6 6 7 7 10 10 8 7 15 15 9 7 15 15 10 7 21 30 30 11 11 21 30 30 12 11 35 42 42 13 13 35 42 42 14 13 35 70 70 15 13 35 105 105 16 13 55 105 105 17 17 55 105 210 210 18 17 77 110 210 210 19 19 77 165 210 210 20 19 91 165 210 210 21 19 91 231 330 330 22 19 91 231 330 330 23 23 91 385 462 462 24 23 143 385 462 462 25 23 143 455 770 770 26 23 143 455 1155 1155 27 23 143 455 1155 1155 28 23 187 455 1365 2310 2310 29 29 187 715 1365 2310 2310 30 29 221 715 1365 2730 2730 31 31 221 1001 1430 2730 2730 32 31 247 1001 2145 2730 2730 33 31 247 1001 2145 2730 2730 34 31 253 1001 3003 4290 4290 35 31 253 1309 3003 4290 4290 36 31 323 1309 5005 6006 6006 37 37 323 1547 5005 6006 6006 38 37 323 1547 5005 10010 10010 39 37 323 1729 5005 15015 15015 40 37 391 1729 6545 15015 15015 41 41 391 2431 6545 15015 30030 30030 42 41 437 2431 7735 15015 30030 30030 43 43 437 2717 7735 19635 30030 30030 44 43 437 2717 8645 19635 30030 30030 45 43 437 2717 8645 23205 39270 39270 46 43 493 2717 12155 23205 39270 39270 47 47 493 3553 12155 25935 46410 46410 48 47 551 3553 17017 25935 46410 46410 49 47 551 4199 17017 36465 51870 51870 50 47 589 4199 19019 36465 51870 51870
